Question

for the reaction 2a + 3b → 4d + 5e, how is the rate of reaction expressed in terms of the rate of disappearance of b? answer: a: +\frac{\deltab}{\delta t} b: -\frac{\deltab}{\delta t} c: +\frac{1}{3}\frac{\deltab}{\delta t} d: -\frac{1}{3}\frac{\deltab}{\delta t} e: +\frac{3}{\deltab\delta t}

Explanation:

Step1: Recall reaction - rate formula

For a general reaction \(aA + bB
ightarrow cC + dD\), the rate of the reaction \(r=-\frac{1}{a}\frac{\Delta[A]}{\Delta t}=-\frac{1}{b}\frac{\Delta[B]}{\Delta t}=\frac{1}{c}\frac{\Delta[C]}{\Delta t}=\frac{1}{d}\frac{\Delta[D]}{\Delta t}\).
For the reaction \(2A + 3B
ightarrow4D+5E\), the rate of the reaction \(r =-\frac{1}{2}\frac{\Delta[A]}{\Delta t}=-\frac{1}{3}\frac{\Delta[B]}{\Delta t}=\frac{1}{4}\frac{\Delta[D]}{\Delta t}=\frac{1}{5}\frac{\Delta[E]}{\Delta t}\).

Step2: Express rate in terms of B

We want to express the rate of the reaction in terms of the rate of disappearance of B. The rate of disappearance of B is \(-\frac{\Delta[B]}{\Delta t}\). The rate of the reaction \(r =-\frac{1}{3}\frac{\Delta[B]}{\Delta t}\), so \(-\frac{\Delta[B]}{\Delta t}=3r\). The rate of the reaction can be written as \(r =-\frac{\Delta[B]}{3\Delta t}\).

Answer:

A. \(\frac{\Delta[B]}{\Delta t}\) is incorrect because the rate of disappearance of B has a negative sign.
B. \(-\frac{\Delta[B]}{\Delta t}\) represents the rate of disappearance of B but not the rate of the reaction in the correct form.
C. \(\frac{1}{3}\frac{\Delta[B]}{\Delta t}\) is incorrect as it has the wrong sign for the reactant - B.
D. \(-\frac{1}{3}\frac{\Delta[B]}{\Delta t}\) is the correct expression for the rate of the reaction in terms of the rate of disappearance of B.
E. \(\frac{3}{\Delta t}\Delta[B]\) is incorrect.
So the answer is D. \(-\frac{1}{3}\frac{\Delta[B]}{\Delta t}\)